Câu 1:

\(4x^2+16x-9\)

\(=4x^2+18x-2x-9\)

\(=2x\left(2x+9\right)-\left(2x+9\right)\)

\(=\left(2x-1\right)\left(2x+9\right)\)

Câu 2:

\(6x^2-11x+3=0\)

\(\Leftrightarrow6x^2-2x-9x+3=0\)

\(\Leftrightarrow2x\left(3x-1\right)-3\left(3x-1\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(3x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=3\\3x=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)

1) \(Q=-x\) khi:

\(\dfrac{x-3}{x+1}=-x\)

\(\Leftrightarrow x-3=-x\left(x+1\right)\)

\(\Leftrightarrow x-3=-x^2-x\)

\(\Leftrightarrow x-3+x^2+x\)

\(\Leftrightarrow x^2+2x-3=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)

2) \(Q< 1\) khi:

\(\dfrac{x-3}{x+1}< 1\)

\(\Leftrightarrow x-3< x+1\)

\(\Leftrightarrow x-x< 1+3\)

\(\Leftrightarrow0< 4\) (luôn đúng) 

Vậy \(Q< 0\) với mọi x 

3) \(Q=m\) khi:

\(\dfrac{x-3}{x+1}=m\)

\(\Leftrightarrow x-3=m\left(x+1\right)\)

\(\Leftrightarrow x-3=mx+m\)

\(\Leftrightarrow x-mx=m+3\)

\(\Leftrightarrow x\left(1-m\right)=m+3\)

\(\Leftrightarrow1-m\ne0\)

\(\Leftrightarrow m\ne1\)

a: XétΔAHB vuông tại H có HM là đường cao

nên BM*BA=BH^2; AM*AB=AH^2; HM*AB=HA*HB

Xét ΔAHC vuông tại H có HN là đường cao

nên AN*AC=AH^2; CN*CA=CH^2; HA*HC=HN*CA

CN*BM*BC

=BH^2/BA*CH^2/CA*BC

\(=\dfrac{\left(BH\cdot CH\right)^2}{BA\cdot CA}\cdot BC\)

=AH^4/AH=AH^3

AM*AB=AH^2

AN*AC=AH^2

=>AM*AB=AN*AC(Cái này mới đúng nè bạn, còn cái AM*AC=AN*AB là sai đề rồi á)

b: AM*AN

=AH^2/AB*AH^2/AC

=AH^4/AB*AC

\(=\dfrac{AH^4}{AH\cdot BC}=\dfrac{AH^3}{BC}\)

c: Sửa đề: AB^3/AC^3=BM/CN

\(\dfrac{BM}{CN}=\dfrac{BH^2}{AB}:\dfrac{CH^2}{AC}\)

\(=\dfrac{BH^2}{AB}\cdot\dfrac{AC}{CH^2}=\dfrac{BH^2}{CH^2}\cdot\dfrac{AC}{AB}=\dfrac{AB^4}{AC^4}\cdot\dfrac{AC}{AB}=\dfrac{AB^3}{AC^3}\)

a: góc C=360-90-60-135=210-135=75 độ

b: Xét ΔABD có AB=AD và góc BAD=60 độ

nên ΔABD đều

a: Xét tứ giác ABMC có

E là trung điểm chung của AM và BC

góc BAC=90 độ

Do đó: ABMC là hình chữ nhật

b: Xét ΔBAC có BD/BA=BE/BC

nên DE//AC

=>EN//AC

Xét tứ giác ANEC có

AN//EC

AC//NE

=>ANEC là hình bình hành

D = (x² - 2xy + y²) + [(2y)²+ 2.2y.1 + 1²] + 2

= (x - y)² + (2y + 1)² + 2

Ta thấy: (x - y)² ≥0∀x thuộc R

              (2y + 1)² ≥0∀y thuộc R

=> (x - y)² + (2y + 1)²  ≥0

=> (x - y)² + (2y + 1)² + 2  ≥2 

=> Min D = 2 \(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\2y-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=y\\y=\dfrac{1}{2}\end{matrix}\right.\)               \(\Rightarrow x=y=\dfrac{1}{2}\)

Vậy Min D = 2 khi x=y=1/2

\(D=x^2+5y^2-2xy+4y+3\)

\(=x^2-2xy+y^2+4y^2+4y+1+2\)

\(=\left(x-y\right)^2+\left(2y+1\right)^2+2\)

Do \(\left(x-y\right)^2\ge0\forall x,y\in R\)

\(\left(2y+1\right)^2\ge0\forall y\in R\)

\(\Rightarrow\left(x-y\right)^2+\left(2y+1\right)^2+2\ge2\forall x,y\in R\)

\(\Rightarrow\) Giá trị nhỏ nhất của D là 2 \(\Leftrightarrow x=y=-\dfrac{1}{2}\)

\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)

\(=\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)-24\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)   (1)

Đặt \(t=x^2+5x+4\)

(1) \(=t\left(t+2\right)-24\)

\(=t^2+2t-24\)

\(=t^2+2t+1-25\)

\(=\left(t+1\right)^2-5^2\)

\(=\left(t+1-5\right)\left(t+1+5\right)\)

\(=\left(t-4\right)\left(t+6\right)\)

\(\Rightarrow\left(1\right)=\left(x^2+5x+4-4\right)\left(x^2+5x+4+6\right)\)

\(=\left(x^2+5x\right)\left(x^2+5x+10\right)\)

\(=x\left(x+5\right)\left(x^2+5x+10\right)\)

Xem lại yêu cầu của đề nhé bạn 

\(A=\dfrac{1}{x^2+x}+\dfrac{1}{x^2+3x+2}+\dfrac{1}{x^2+5x+6}+\dfrac{1}{x+3}\) (ĐK: \(x\ne-1;x\ne0;x\ne-2;x\ne-3\))

\(A=\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{x+3}\)

\(A=\dfrac{\left(x+2\right)\left(x+3\right)}{x\left(x+1\right)\left(x+2\right)\left(x+3\right)}+\dfrac{x\left(x+3\right)}{x\left(x+1\right)\left(x+2\right)\left(x+3\right)}+\dfrac{x\left(x+1\right)}{x\left(x+1\right)\left(x+2\right)\left(x+3\right)}+\dfrac{x\left(x+1\right)\left(x+2\right)}{x\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)\(A=\dfrac{x^2+5x+6+x^2+3x+x^2+x+x^3+2x^2+x^2+2x}{x\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)

\(A=\dfrac{x^3+6x^2+11x+6}{x\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)

\(A=\dfrac{x^3+5x^2+6x+x^2+5x+6}{x\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)

\(A=\dfrac{x\left(x+5x+6\right)+\left(x^2+5x+6\right)}{x\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)

\(A=\dfrac{\left(x^2+5x+6\right)\left(x+1\right)}{x\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)

\(A=\dfrac{\left(x+1\right)\left(x+2\right)\left(x+3\right)}{x\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)

\(A=\dfrac{1}{x}\)

`@` `\text {Ans}`

`\downarrow`

\(B=(x+1)^2-2(2x-1)(1+x)+4x^2-4x+1\)

`= x^2 + 2x + 1 - 2(2x^2 + x - 1) + 4x^2 - 4x + 1`

`= 5x^2 - 2x + 2 - 4x^2 - 2x + 2`

`= x^2 - 4x + 4`

\(B=\left(x+1\right)^2-2\left(2x-1\right)\left(1+x\right)+4x^2-4x+1\)

\(=\left(x+1\right)^2-2\left(x+1\right)\left(2x-1\right)+\left(2x-1\right)^2\)

\(=\left(x+1-2x+1\right)^2\)

\(=\left(2-x\right)^2\)

\(2x^2-x-8=0\\ \Leftrightarrow\left(2x^2-x\right)-8=0\\ \Leftrightarrow x\left(2x-1\right)-8=0\\ \Leftrightarrow\left(x-8\right)\left(2x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x=\dfrac{1}{2}\end{matrix}\right.\)

x(2x-1)-8 sao lại bằng (x-8)(2x-1) được ạ